ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
jHbuu287uxc-036\|And there's kind of an analogy.
jHbuu287uxc-037\|Think about it-- if you were to sneeze in a crowded room, a hot, noisy room, that sneeze wouldn't have much effect.
jHbuu287uxc-038\|But think about that same sneeze in an absolutely quiet, cool room.
jHbuu287uxc-039\|Then there'd be a big effect of that sneeze.
jHbuu287uxc-040\|So you're seeing that same thing.
jHbuu287uxc-041\|The same amount of heat transferring have a bigger effect on the entropy on the cold side than the hot side.
jHbuu287uxc-042\|So if you add those two, you find that the overall entropy increases.
jHbuu287uxc-043\|Now, what if the temperature changes?
jHbuu287uxc-049\|So the temperature change and the heat capacity determine the entropy change if the temperature changes.
jHbuu287uxc-050\|In fact, it's very strongly related to the heat capacity of the system.
jHbuu287uxc-051\|It determines how the entropy changes in the system.
jHbuu287uxc-053\|Delta-S, q-reversible over T gives us a way to measure heat and calculate entropy.
jHbuu287uxc-054\|Much easier than taking microstates and trying to count them.
jHbuu287uxc-055\|The statistical and thermodynamic versions of entropy, turns out the thermodynamic one-- much easier to measure.
6EOSsIa9wDg-000\|In oxidation reduction chemistry, we tabulate standard reduction potentials for half cells.
6EOSsIa9wDg-001\|And we tabulate them relative to the standard hydrogen electrode.
6EOSsIa9wDg-002\|We arbitrarily set that electrode to zero, and we measure all our potentials relative to that one.
6EOSsIa9wDg-006\|We can look at the top one here is permanganate being reduced.
6EOSsIa9wDg-007\|Now, when permanganate is reduced, that has a high reduction potential.
6EOSsIa9wDg-010\|So as it is being reduced, as this reduction occurs, the permanganate is oxidizing something else.
6EOSsIa9wDg-011\|That is, when you pull electrons off and become reduced, you had to oxidize something.
6EOSsIa9wDg-012\|The process of pulling electrons off something is an oxidation.
6EOSsIa9wDg-013\|So high reduction potential means you have a strong ability to oxidize.
6EOSsIa9wDg-014\|So the permanganate ion can oxidize any of the products here listed below.
6EOSsIa9wDg-015\|And in fact, all these products and the products in the lower half of our table as well.
6EOSsIa9wDg-020\|Let's look at the lower half of the table.
6EOSsIa9wDg-021\|We can also look at those reduction potentials lower than the standard hydrogen electrode.
6EOSsIa9wDg-024\|In contrast, it means calcium metal is a strong reducing agent.
6EOSsIa9wDg-025\|That is the preferred direction is essentially the reverse direction, to give up electrons and form calcium ions.
6EOSsIa9wDg-026\|If you give up your electrons, something else must accept them, and that thing that accepts them becomes reduced.
6EOSsIa9wDg-027\|Its oxidation number is reduced by accepting negatively charged electrons.
3ofEcJm3U7Y-000\|Now let's see how good your memory is.
3ofEcJm3U7Y-001\|We're going to look at those three atoms and see if we can choose which will have this ionization profile.
3ofEcJm3U7Y-002\|Remember, when you ionize an electron, if you give excess photon energy, that will go into kinetic energy of the ejected electron.
3ofEcJm3U7Y-011\|So we're testing your memory a little bit.
3ofEcJm3U7Y-013\|You can have some transitions in the visible, but if you're going to ionize, that's going to be in the ultraviolet.
3ofEcJm3U7Y-014\|And among these three, we said hydrogen will be the most difficult to ionize.
3ofEcJm3U7Y-015\|They'll all behave exactly like hydrogen because we saw that their effective nuclear charges are about 1.
3ofEcJm3U7Y-019\|So they would come here at lower frequency thresholds than the ultraviolet.
yMvYRVCIEfo-000\|One situation where the thermodynamic definition of entropy that is the heat evolved at a constant temperature is particularly appealing our phase changes.
yMvYRVCIEfo-001\|That's because phase changes occur at a fixed temperature, and there's a specific heat involved, the enthalpy of either vaporization or melting.
yMvYRVCIEfo-003\|And you might expect that regardless of the liquid that this entropy of vaporization is going to be about the same.
yMvYRVCIEfo-005\|So you're going to expect an increase in entropy, and it will be fairly independent of the nature of the particle.
yMvYRVCIEfo-006\|And that's what you find.
yMvYRVCIEfo-007\|If you have a higher enthalpy of vaporization you boil at a higher temperature.
UNmXHX7WfWs-000\|Now let's talk about electrons grouping around a nucleus.
UNmXHX7WfWs-001\|So they have to occupy many of the orbitals, and we've already seen when they do that.
UNmXHX7WfWs-002\|They occupy the orbitals in regular ways.
UNmXHX7WfWs-003\|There's rules that say, no two electrons can have the same four quantum numbers.
UNmXHX7WfWs-004\|N l m sub l and m sub s, at least one must be different for each electron.
UNmXHX7WfWs-005\|And we've seen they always go in spin parallel to degenerate orbitals.
UNmXHX7WfWs-006\|But another effect occurs, and that is low energy electrons will shield the higher energy electrons.
UNmXHX7WfWs-008\|You can't solve it exactly like you can the two body problem, one nucleus and one electron.
UNmXHX7WfWs-009\|So what we do is we take the one nucleus one electron problem, and we say, let that one electron define all the orbitals.
UNmXHX7WfWs-010\|And then when we start putting more electrons in, they'll serve as just a slight perturbation to that one electron system.
UNmXHX7WfWs-011\|It's easier than solving all the mathematics over again.
UNmXHX7WfWs-013\|Now, s electrons are very good shielders.
UNmXHX7WfWs-014\|That is, they can shield some of the nuclear charge from outer electrons.
UNmXHX7WfWs-015\|So here's a big nuclear positive charge.
UNmXHX7WfWs-016\|An s electron has no node at the nucleus.
UNmXHX7WfWs-017\|That is, the orbitals are aligned such that the electron has access to the nucleus.
UNmXHX7WfWs-018\|If you're a p electron or a d electron, if l is one or two, then you have an angular node at the nucleus.
UNmXHX7WfWs-019\|You're forbidden mathematically to exist at the nucleus.
UNmXHX7WfWs-022\|Now, what do we mean by shielding?
UNmXHX7WfWs-023\|Shielding means I remove some of the effective nuclear charge.
UNmXHX7WfWs-025\|And it'll seem more like maybe just a plus one-ish effective nuclear charge.
UNmXHX7WfWs-026\|So if it sees a lower nuclear charge, it's easier to ionize.
UNmXHX7WfWs-027\|Its energy state is raised, it's closer to that zero ionized state.
UNmXHX7WfWs-028\|Let's look at that in more detail.
UNmXHX7WfWs-029\|Here I've plotted the one s, two s, three s, four s energy levels.
UNmXHX7WfWs-030\|L equals zero.
UNmXHX7WfWs-031\|Very good shielding electrons, because l equals 0 means s, no node at the nucleus.
UNmXHX7WfWs-033\|These levels are raised slightly because of the shielding effect of the s electron.
UNmXHX7WfWs-034\|And that effect continues.
UNmXHX7WfWs-036\|And the effect is pretty dramatic.
UNmXHX7WfWs-039\|S electrons that have this inner state don't shield each other very well.
UNmXHX7WfWs-040\|All s electrons have equal access to the nucleus.
UNmXHX7WfWs-042\|But s electrons are great shielders of p and d electrons.
UNmXHX7WfWs-043\|So you get two p energies that are higher than 2 s, and you get d energies that are even higher than four s.
idWS5CF3e8U-000\|We measure heat transfer using a technique called "calorimetry." Calorimetry revolves around the conservation of energy.
idWS5CF3e8U-001\|That is, every joule of heat lost by a system is absorbed by the surroundings.
idWS5CF3e8U-002\|Or, every joule of heat absorbed by the system comes from the surroundings.
idWS5CF3e8U-007\|Now, what causes heat capacities?
idWS5CF3e8U-008\|Well, if you add heat to a system and raise its temperature, what you're doing is changing the kinetic energies of the particles.
idWS5CF3e8U-009\|You can always associate higher temperatures with more molecular motion.
idWS5CF3e8U-010\|So, higher temperature, and I've absorbed some heat, means that heat went into increasing the kinetic energy of those particles.
idWS5CF3e8U-015\|More molecular motion is higher temperature.
idWS5CF3e8U-018\|That is, I'm getting more motion for every joule of heat that I add.
idWS5CF3e8U-019\|Well, what if I don't get more relative motion?
idWS5CF3e8U-023\|I completely overcome the hydrogen bonding interactions in liquid water and add energy to do that transition.
idWS5CF3e8U-024\|Now, there I'm adding energy, and I'm getting a phase transition, rather than an increase in kinetic energy.
idWS5CF3e8U-025\|That's a potential energy contribution to the heat capacity.
idWS5CF3e8U-026\|Or I could have intramolecular bonds breaking or forming.
W_ngfrFO7f0-000\|Let's look at a sequential series of reactions, A going to B going to C.
W_ngfrFO7f0-008\|We're talking about a 2-step reaction.
W_ngfrFO7f0-009\|A goes to B and then B goes to C.
W_ngfrFO7f0-010\|The enthalpy for the first step, delta H1, the enthalpy for the second step, delta H2.
W_ngfrFO7f0-013\|That is, I put in energy here to break those bonds but then I get all of that back because I form the same bonds.
W_ngfrFO7f0-014\|So the sequential reaction A going to C, delta H1 plus delta H2 would be the enthalpy for that reaction.
W_ngfrFO7f0-015\|Those are equal and opposite.

jHbuu287uxc-036|And there's kind of an analogy.

jHbuu287uxc-037|Think about it-- if you were to sneeze in a crowded room, a hot, noisy room, that sneeze wouldn't have much effect.

jHbuu287uxc-038|But think about that same sneeze in an absolutely quiet, cool room.

jHbuu287uxc-039|Then there'd be a big effect of that sneeze.

jHbuu287uxc-040|So you're seeing that same thing.

jHbuu287uxc-041|The same amount of heat transferring have a bigger effect on the entropy on the cold side than the hot side.

jHbuu287uxc-042|So if you add those two, you find that the overall entropy increases.

jHbuu287uxc-043|Now, what if the temperature changes?

jHbuu287uxc-049|So the temperature change and the heat capacity determine the entropy change if the temperature changes.

jHbuu287uxc-050|In fact, it's very strongly related to the heat capacity of the system.

jHbuu287uxc-051|It determines how the entropy changes in the system.

jHbuu287uxc-053|Delta-S, q-reversible over T gives us a way to measure heat and calculate entropy.

jHbuu287uxc-054|Much easier than taking microstates and trying to count them.

jHbuu287uxc-055|The statistical and thermodynamic versions of entropy, turns out the thermodynamic one-- much easier to measure.

6EOSsIa9wDg-000|In oxidation reduction chemistry, we tabulate standard reduction potentials for half cells.

6EOSsIa9wDg-001|And we tabulate them relative to the standard hydrogen electrode.

6EOSsIa9wDg-002|We arbitrarily set that electrode to zero, and we measure all our potentials relative to that one.

6EOSsIa9wDg-006|We can look at the top one here is permanganate being reduced.

6EOSsIa9wDg-007|Now, when permanganate is reduced, that has a high reduction potential.

6EOSsIa9wDg-010|So as it is being reduced, as this reduction occurs, the permanganate is oxidizing something else.

6EOSsIa9wDg-011|That is, when you pull electrons off and become reduced, you had to oxidize something.

6EOSsIa9wDg-012|The process of pulling electrons off something is an oxidation.

6EOSsIa9wDg-013|So high reduction potential means you have a strong ability to oxidize.

6EOSsIa9wDg-014|So the permanganate ion can oxidize any of the products here listed below.

6EOSsIa9wDg-015|And in fact, all these products and the products in the lower half of our table as well.

6EOSsIa9wDg-020|Let's look at the lower half of the table.

6EOSsIa9wDg-021|We can also look at those reduction potentials lower than the standard hydrogen electrode.

6EOSsIa9wDg-024|In contrast, it means calcium metal is a strong reducing agent.

6EOSsIa9wDg-025|That is the preferred direction is essentially the reverse direction, to give up electrons and form calcium ions.

6EOSsIa9wDg-026|If you give up your electrons, something else must accept them, and that thing that accepts them becomes reduced.

6EOSsIa9wDg-027|Its oxidation number is reduced by accepting negatively charged electrons.

3ofEcJm3U7Y-000|Now let's see how good your memory is.

3ofEcJm3U7Y-001|We're going to look at those three atoms and see if we can choose which will have this ionization profile.

3ofEcJm3U7Y-002|Remember, when you ionize an electron, if you give excess photon energy, that will go into kinetic energy of the ejected electron.

3ofEcJm3U7Y-011|So we're testing your memory a little bit.

3ofEcJm3U7Y-013|You can have some transitions in the visible, but if you're going to ionize, that's going to be in the ultraviolet.

3ofEcJm3U7Y-014|And among these three, we said hydrogen will be the most difficult to ionize.

3ofEcJm3U7Y-015|They'll all behave exactly like hydrogen because we saw that their effective nuclear charges are about 1.

3ofEcJm3U7Y-019|So they would come here at lower frequency thresholds than the ultraviolet.

yMvYRVCIEfo-000|One situation where the thermodynamic definition of entropy that is the heat evolved at a constant temperature is particularly appealing our phase changes.

yMvYRVCIEfo-001|That's because phase changes occur at a fixed temperature, and there's a specific heat involved, the enthalpy of either vaporization or melting.

yMvYRVCIEfo-003|And you might expect that regardless of the liquid that this entropy of vaporization is going to be about the same.

yMvYRVCIEfo-005|So you're going to expect an increase in entropy, and it will be fairly independent of the nature of the particle.

yMvYRVCIEfo-006|And that's what you find.

yMvYRVCIEfo-007|If you have a higher enthalpy of vaporization you boil at a higher temperature.

UNmXHX7WfWs-000|Now let's talk about electrons grouping around a nucleus.

UNmXHX7WfWs-001|So they have to occupy many of the orbitals, and we've already seen when they do that.

UNmXHX7WfWs-002|They occupy the orbitals in regular ways.

UNmXHX7WfWs-003|There's rules that say, no two electrons can have the same four quantum numbers.

UNmXHX7WfWs-004|N l m sub l and m sub s, at least one must be different for each electron.

UNmXHX7WfWs-005|And we've seen they always go in spin parallel to degenerate orbitals.

UNmXHX7WfWs-006|But another effect occurs, and that is low energy electrons will shield the higher energy electrons.

UNmXHX7WfWs-008|You can't solve it exactly like you can the two body problem, one nucleus and one electron.

UNmXHX7WfWs-009|So what we do is we take the one nucleus one electron problem, and we say, let that one electron define all the orbitals.

UNmXHX7WfWs-010|And then when we start putting more electrons in, they'll serve as just a slight perturbation to that one electron system.

UNmXHX7WfWs-011|It's easier than solving all the mathematics over again.

UNmXHX7WfWs-013|Now, s electrons are very good shielders.

UNmXHX7WfWs-014|That is, they can shield some of the nuclear charge from outer electrons.

UNmXHX7WfWs-015|So here's a big nuclear positive charge.

UNmXHX7WfWs-016|An s electron has no node at the nucleus.

UNmXHX7WfWs-017|That is, the orbitals are aligned such that the electron has access to the nucleus.

UNmXHX7WfWs-018|If you're a p electron or a d electron, if l is one or two, then you have an angular node at the nucleus.

UNmXHX7WfWs-019|You're forbidden mathematically to exist at the nucleus.

UNmXHX7WfWs-022|Now, what do we mean by shielding?

UNmXHX7WfWs-023|Shielding means I remove some of the effective nuclear charge.

UNmXHX7WfWs-025|And it'll seem more like maybe just a plus one-ish effective nuclear charge.

UNmXHX7WfWs-026|So if it sees a lower nuclear charge, it's easier to ionize.

UNmXHX7WfWs-027|Its energy state is raised, it's closer to that zero ionized state.

UNmXHX7WfWs-028|Let's look at that in more detail.

UNmXHX7WfWs-029|Here I've plotted the one s, two s, three s, four s energy levels.

UNmXHX7WfWs-030|L equals zero.

UNmXHX7WfWs-031|Very good shielding electrons, because l equals 0 means s, no node at the nucleus.

UNmXHX7WfWs-033|These levels are raised slightly because of the shielding effect of the s electron.

UNmXHX7WfWs-034|And that effect continues.

UNmXHX7WfWs-036|And the effect is pretty dramatic.

UNmXHX7WfWs-039|S electrons that have this inner state don't shield each other very well.

UNmXHX7WfWs-040|All s electrons have equal access to the nucleus.

UNmXHX7WfWs-042|But s electrons are great shielders of p and d electrons.

UNmXHX7WfWs-043|So you get two p energies that are higher than 2 s, and you get d energies that are even higher than four s.

idWS5CF3e8U-000|We measure heat transfer using a technique called "calorimetry." Calorimetry revolves around the conservation of energy.

idWS5CF3e8U-001|That is, every joule of heat lost by a system is absorbed by the surroundings.

idWS5CF3e8U-002|Or, every joule of heat absorbed by the system comes from the surroundings.

idWS5CF3e8U-007|Now, what causes heat capacities?

idWS5CF3e8U-008|Well, if you add heat to a system and raise its temperature, what you're doing is changing the kinetic energies of the particles.

idWS5CF3e8U-009|You can always associate higher temperatures with more molecular motion.

idWS5CF3e8U-010|So, higher temperature, and I've absorbed some heat, means that heat went into increasing the kinetic energy of those particles.

idWS5CF3e8U-015|More molecular motion is higher temperature.

idWS5CF3e8U-018|That is, I'm getting more motion for every joule of heat that I add.

idWS5CF3e8U-019|Well, what if I don't get more relative motion?

idWS5CF3e8U-023|I completely overcome the hydrogen bonding interactions in liquid water and add energy to do that transition.

idWS5CF3e8U-024|Now, there I'm adding energy, and I'm getting a phase transition, rather than an increase in kinetic energy.

idWS5CF3e8U-025|That's a potential energy contribution to the heat capacity.

idWS5CF3e8U-026|Or I could have intramolecular bonds breaking or forming.

W_ngfrFO7f0-000|Let's look at a sequential series of reactions, A going to B going to C.

W_ngfrFO7f0-008|We're talking about a 2-step reaction.

W_ngfrFO7f0-009|A goes to B and then B goes to C.

W_ngfrFO7f0-010|The enthalpy for the first step, delta H1, the enthalpy for the second step, delta H2.

W_ngfrFO7f0-013|That is, I put in energy here to break those bonds but then I get all of that back because I form the same bonds.

W_ngfrFO7f0-014|So the sequential reaction A going to C, delta H1 plus delta H2 would be the enthalpy for that reaction.

W_ngfrFO7f0-015|Those are equal and opposite.